Practice Questions for Midterm 2 - Math 1060Q - Fall 2013

Transcription

1 Eam Review Practice Questions for Midterm - Math 060Q - Fall 0 The following is a selection of problems to help prepare ou for the second midterm eam. Please note the following: anthing from Module/Chapter is fair game. That is, anthing related to trigonometr could be on there. there ma be mistakes steven.pon@uconn.edu if ou find one. the distribution of problems on this review sheet does not reflect the distribution that will be on the eam. learning math is about more than just memorizing the steps to certain problems. You have to understand the concepts behind the math. In order to test our understanding of the math, ou ma see questions on the eam that are unfamiliar, but that rel on the concepts ou ve learned. Therefore, ou should concentrate on learning the underling theor, not on memorizing steps without knowing wh ou re doing each step. (Some eamples of unfamiliar problems can be found in the sample problems below, to give ou an idea.) First, a few questions on graphing rational functions.. Given the function f() = 9 ( ) : (a) Find an vertical asmptotes of f. (b) Find an horizontal asmptotes of f. (c) Find an zeros of f. (d) What is the domain of f? (e) Sketch a graph of f. Note that we can simplif f as follows: f() = (a) f has a vertical asmptote at = 0 (b) f has a horizontal asmptote at = (c) f has a zero at = ( )( + ) ( ) = +. (d) The domain of f is (, 0) (0, ) (, ) (i.e., f has a hole at ) (e) f looks like: Page of

2 Eam Review. Given the function f() = ( + )( ) ( + )( + )( + 7) : (a) Find an vertical asmptotes of f. (b) Find an horizontal asmptotes of f. (c) Find an zeros of f. (d) What is the domain of f? (e) Sketch a graph of f. Note that we can simplif f as follows: f() = ( + )( + 7). (a) f has a vertical asmptote at = and = 7. (b) f has a horizontal asmptote at = 0 (c) f has a zero at = (d) The domain of f is (, 7) ( 7, ) (, ) (, ) (e) f looks like: Graph the function f() = + +, clearl indicating an zeros, holes, and asmptotes. In this case, we can rewrite f as: f() = + ( + ) + ( 6) ( ) = = + ( )( + ) ( )( + ). Therefore, f has no holes, has a vertical asmptote at = and =, has a horizontal asmptote at = 0, and has a zero at =. A little more work plugging in numbers figures out which wa the function goes at each asmptote (to positive or negative infinit). Page of

3 Eam Review Now trig functions. Angles can be measured in degrees or radians, and though we primaril use radians, it s worth knowing how to convert. More importantl, ou should know the unit circle ver well!. Convert 0 to radians. 0 = 6 radians 5. Convert radians to degrees. = Which of the following angles correspond to the same spot on the unit circle?, 5, 9, 7, 9 and 7 all correspond. To see this add + = 9. On the other hand, if we substract = What is an angle in the range [, ] that corresponds to the angle 6 6 )? (i.e., is coterminal with We can find the answer b the addition of since our starting angle is below our desired range. Notice that 6 + = What is an angle in the range [, 0] that corresponds to the angle 5 5 )? (i.e., is coterminal with Just as above, 5 the desired range. =. In this case we substract in order to move into 9. Draw the unit circle. Label the angles 0, 6,,,, 7 6, 7,,, 5. Label the coordinates of the points on the unit circle that correspond to those angles. 0. If a point P (t) on the unit circle has coordinates ( /5, /5), then what are the coordinates of: Page of

4 Eam Review (a) P (t + ) (b) P (t + ) (c) P ( ) (d) P (t + 00) (e) P ( t) Note: ou don t need to find what t is! (a) P (t + ) For this problem and all of the others we will use the sum of two angles identities: cos(a ± b) = cos(a) cos(b) sin(a) sin(b) sin(a ± b) = sin(a) cos(b) ± cos(a) sin(b) to figure out how each point changes. Notice that sin(t) = 5 and cos(t) = 5. So P (t + )=(new, new ) can be calculated b finding each of the new coordinates. new = cos(t + ) = cos(t) cos() sin(t) sin() = cos(t)( ) sin(t)(0) = 5 new = sin(t + ) = sin(t) cos() + cos(t) sin() = sin(t)( ) + cos(t)(0) = 5 So P (t + ) = ( 5, 5 ). (b) P (t + ) One can use the same process for this problem and all of the others. new = cos(t + ) = cos(t) cos( ) sin(t) sin( ) = cos(t)(0) sin(t)() = 5 new = sin(t + ) = sin(t) cos( ) + cos(t) sin( ) = sin(t)(0) + cos(t)() = 5 So P (t + ) = ( 5, 5 ). (c) P ( ) Here cos( ) = 0 while sin( ) = so P ( ) = (0, ). (d) P (t + 00) One could use the same method as in part a) so instead consider using coterminal angles to evaluate P (t+00). First, notice that 00 is eactl 50 rotations around the unit circle. So, both the and coordinates will be changed. Therefore P (t+00) = P (t) = ( 5, 5 ) (e) P ( t) One could use the same method as in the first problem here as well, so for the sake of variet consider solving this problem using the fact that cosine is an even function, and sine is an odd function. Recall that a function is odd if it satisfies the equation: f() = f( ) while a function is even if it satisfies the equation: f() = f( ) for all in its domain. Using the above one concludes that cos( t) = cos(t) and sin( t) = sin(t). Therefore, P ( t) = ( 5, 5 ).. Find an angle in [0, ) that is coterminal with the angle Since 5 6 = = is coterminal with the angle Find an angle in [0, ) that is coterminal with the angle 5. Since 5 = = 5 + is coterminal with the angle Find the reference angle for the angle 7. (The reference angle is the angle in [0, ] formed between a given angle and the -ais, so for eample, the reference angle for is.) Page of

5 Eam Review The reference angle for 7 is.. Find the reference angle for the angle 5 6. The reference angle for 5 6 is Find the reference angle for the angle 7 5. The reference angle for 7 5 is 5. Do ou understand what the trigonometric functions are? both using triangles or the unit circle? How to calculate them, 6. Define sine, cosine, tangent, cosecant, secant, cotangent. What are their domain and range? Let θ be the angle formed starting with the positive -ais and going counterclockwise if θ is positive and clockwise if θ is negative. Then the corresponding terminal point on the unit circle has coordinates (cos(θ), sin(θ)). The other trig functions are defined as follows: tan(θ) = sin(θ) cos(θ), cot(θ) = cos(θ) sin(θ), sec(θ) = cos(θ), csc(θ) = ) sin(θ) Function Domain Range sin(θ) (, ) [, ] cos(θ) (, ) [, ] tan(θ) all real numbers ecept n for odd integers n (, ) cot(θ) all real numbers ecept n for all integers n (, ) sec(θ) all real numbers ecept n for odd integers n (, ] [, ) csc(θ) all real numbers ecept n for all integers n (, ] [, ) For angles between 0 and, these trig functions can be equivalentl defined using right triangles. In a right triangle, sin(θ) = opposite hpotenus tan(θ) = opposite adjacent sec(θ) = hpotenuse adjacent 7. What is sin(0 )?, cos(θ) = adjacent hpotenus, cot(θ) = adjacent opposite, csc(θ) = hpotenuse opposite. sin(0 ) = 8. What is csc( 5 6 )? csc( 5 6 ) = sin( 5 6 ) = = 9. If t = 0, what are sin(t), csc(t), and cot(t)? sin(t) = csc(t) = sin(t) = cot(t) = cos(t) sin(t) = = = 0. If cot(t) = and t is in the interval [, ], then what is sin(t)? Page 5 of

6 Eam Review Since cot(t) = cos(t) sin(t) = t = 5 is the angle in [, ] where cot(t) =. sin(t) = sin( 5 ) =. If cos(t) = and t is in the interval [, ], then what is tan(t)? t = is the angle in [, ] where cos(t) =. tan(t) = sin(t) cos(t) = =. Given that cos(θ) = 7 and θ is in quadrant IV, find sin(θ). Opposite side of θ is ± 7 = ± 5 Since sin(θ) is negative in quadrant III, so sin(θ) = 5 7. Given that tan(θ) = 5 and θ is in quadrant II, find csc(θ). Hpotenuse side of θ is + ( 5) = so csc(θ) = sin(θ) = =. Use the triangle below to answer these questions. (a) Sa α = and BC = 8. What is AB? (b) Sa β = 6 and BC = 0. What is AC? (c) Sa AB = 5 an BC = 0. What is α? (a) tan( ) = 8 = Therefore, = 8 (b) cos( 6 ) = 0 So = 0 = (c) tan(α) = 0 5 = So α = arctan() Page 6 of

7 Eam Review What about going backwards? If ou know the values of trig functions, can ou work backwards to find out what the angle might be? 5. If sin(t) =, what might t be? Give all possible solutions in [, ]. t =, 6. If sin t =, what might t be? Give all possible solutions in [0, ]. t = 7. If cos t =, what might t be? Give all possible solutions in [0, ]. t =, 5 8. If tan t =, what might t be? Give all possible solutions in [0, ]. t =, 7 9. Find all values of t in the interval [0, ] such that sin t = 0. t = 0,, 0. Find all values of t in the interval [0, ] such that csc t =. i.e., sin(t) = so sin(t) = and t = 6, Find all values of t in the interval [0, ] such that cot t =. i.e., tan(t) =, so tan(t) =. Like an other function, we sometimes want to graph trigonometric functions. Can ou? (We won t worr about reall complicated graphs with lots of transformations, like 5 sin( 6 ). But simpler ones, like below, ou could have to graph.. Graph sin(), cos(), and tan(). sin() Page 7 of

8 Eam Review cos() tan() Graph f() = sin(). sin(). Graph g() = cos( ). Can ou find another function that has this same graph? Page 8 of

9 Eam Review cos( ) This is the same as sin(). 5. Graph h() = sin(). sin( ) 6. Where does csc() have a vertical asmptote? The function csc() is the same as sin(), so it is undefined (and has a vertical asmptote) whenever sin() = 0. That s at multiples of : 0,,, etc. 7. Graph sec(). Page 9 of

10 Eam Review sec() Write an equation for this graph. This could be a sine or cosine graph. Since it starts at zero, it is probabl easiest to write it as a modified sine graph. Note that the amplitude is, so it will be something like sin(stuff). Then note that the period is, so this graph has been stretched horizontall b a factor of. Therefore, the graph is sin( ). A big thing needed in calculus is solving trigonometric equations, or in other words, solving equations that just happen to involve trigonometric functions. Here are a bunch to work on. 9. Find all solutions to sin() + = 0 in [0, ). sin() + = 0 0. Find all solutions to cos = in [0, ]. sin() = = 7 6, 6 Page 0 of

11 Eam Review cos = cos = = 0,. Find all solutions to sin = sin in [0, ]. sin = sin sin = sin = =,. Find all solutions to cos = 0 in [0, ]. cos = cos = cos = ± =,,, 5. Find all solutions to cos() = in [0, ]. cos() =. Find all solutions to tan () = in [0, ]. =, 5, 7, = 6, 5 6, 7 6, 6 tan () = tan() = =,,, 5 5. Find all solutions to sin + = sin in [, ]. Page of

12 Eam Review sin + = sin sin = sin = = 5, 7 =, 6. Find all solutions to cos( ) = 0 in [0, ]. cos( ) = 0 cos( ) = 0 =,, 5,... = + 6, +, Find all solutions to sec = cos in [0, ]. cos = cos = cos = cos cos = ± =,, 5, 7 8. Find all solutions to sin sin = 0 in [0, ]. sin sin = 0 ( sin + )(sin ) = 0 sin =, =, 7 6, 6 9. Find all solutions to sin cos + cos = 0 in [0, ]. Page of

13 Eam Review sin cos + cos = 0 cos (sin ) = 0 cos = 0 or sin = =, 50. Find all solutions to sin() = cos() in [0, ]. sin() = cos() sin cos = tan = =, 7,, 5,... = 8, 7 8, 8, 5 8 Sometimes ou need to use a trigonometric identit to help solve an equation. You ll need to use trig identities in the following problems. A hint is given for each one. 5. Find all solutions to cos + sin = in [0, ). cos + sin = ( sin ) + sin = 0 sin sin + = 0 ( sin ) (sin ) = 0 sin = or sin = = 6,, Find all solutions to sin() = cos in [0, ). sin() = cos sin cos = cos sin cos cos = 0 ( cos sin ) = 0 cos = 0 or sin = =,,, Page of

14 Eam Review 5. Find all solutions to sin(t) tan(t) = 0 in [0, ). sin(t) tan(t) = 0 sin(t) cos(t) sin(t) = 0 ( sin(t) ) cos(t) = 0 sin(t) = 0 or cos(t) = t = 0,,,,... or t =, 7, 9, 5,... t = 0,,,, 8, 7 8, 9 8, Find all solutions to cot + csc = 0 in [0, ]. cot + csc = 0 cot + ( + cos ) = 0 cot = 0 cot = ± =,,, Find all solutions to cos() = cos in [0, ]. cos() = cos cos = cos cos cos = 0 ( cos + ) (cos ) = 0 cos = or cos = = 0,, Just like with other functions we ve studied, trigonometric functions have inverses. Well, partial inverses (since trig functions aren t one-to-one). Do ou know what the inverse trig functions are? Do ou know their domain and range? Can ou compute them for some values? Can ou graph them? 56. What is arccos? θ = Page of

15 Eam Review 57. What is arcsin? θ = 58. What is arctan 0? Let arctan 0=θ. Note: θ must be in ( /, /). tan(θ) = 0. Therefore, θ = What is sin (sin( ))? Let sin (sin( ))=θ. Note: θ must be in [ /, /]. since is coterminal with sin (sin( ))=sin (sin( Answer: θ = 60. What is tan(arctan( ))? ))= Note: tan(arctan()) = if is in (-infinit,infinit). Since - is in (-infinit,infinit) Answer: tan(arctan( ))=- 6. What is arccos(cos( 9 ))? Note: arccos(cos(θ)) = θ if θ is in the range of the arccosine function, i.e. [0, ]. Since cosine function is an even function cos( 9 ) = cos( 9 ) arccos(cos( 9 ))=arccos(cos( 9 ))= 9 Answer: 9 6. What is tan(sin ( )? Let sin ( )=θ. Then sin θ=. Using SOH,CAH,TOA O=, H=, A= and tan θ= = So tan(sin ( )=tan θ= Answer: 6. What is tan(arccos( 5 ))? Let arccos( 5 )=θ. Then cos θ= 5. Using SOH,CAH,TOA A=5, H=, O= and tan θ= 5 So tan(arccos( 5 Answer: 5 )=tan θ= 5 Page 5 of

16 = tan () Math 060 Eam Review 6. Epress a solution to the equation tan( ) = 5 using inverse trigonometric functions. arctan(5) = = arctan(5) Epress a solution to the equation sin() + = using inverse trigonometric functions. sin() = = arcsin( ) = arcsin( ) 66. Graph the function cos (). = arccos() 67. Graph the function arctan(). 68. Graph the functions csc () and csc(). Page 6 of

17 Eam Review = csc () = csc() 0.5 Note that csc() = sin() = sin(). Can ou appl our knowledge to real-world applications? 69. How long a ladder do ou need if ou want to reach a window that s 0 feet off the ground? You re on uneven terrain, so leaning a ladder an steeper than 0 degrees is unsafe. The ladder would be the hpotenuse of a triangle that has height 0 feet and is leaned at an angle of 0 degrees (note that this is a completel unrealistic situation). Then sin(0 ) = 0, where is the length of the ladder. Therefore, = 0, so = An equilateral triangle is circumscribed around a circle of radius 5 (thus, the center of the triangle and the center of the circle are at the same point). What is the area of the triangle? Equilateral triangles have angles of size. If the triangle is circumscribed around the circle, that means the distance from the center of the triangle to one of its edges is the same as the radius of the circle, so it s 5. Then ou can make a right triangle, with a line connecting the center of the circle to the edge of the triangle and another line connecting the center of the circle to a verte of the triangle. This cuts one of the angles of the equilateral triangle in half, so the right triangle has an angle of 6. The opposite side to that angle has length 5, so sin( 6 ) = 5, or in other words, = 0 is the length of the line connecting the center of the circle to a verte of the triangle. Therefore, the height of the equilateral triangle is 5 (draw a picture to help see wh that s the case). Then, let s find the length of one side of the equilateral triangle using cos( 6 ) = z 0, where z is half the length of one side of the Page 7 of

18 Eam Review equilateral triangle. That means that z = 5, so the length of the base of the equilateral triangle is 0. Therefore, the area of the triangle is Car A drives 0 miles north and then 8 miles east. Car B drives 0 miles north, and then miles west. The start from the same point. (a) What is the distance between the cars? (b) Let s sa the starting point of both cars is the origin, and east is the -ais, north is the -ais, etc. What is the angle between Car A and the -ais? (Epressed in radians or using inverse trig functions if necessar.) What is the angle between the two cars? (a) Set up a picture of the situation. We can etrapolate a right triangle with side lengths and 0 whose hpotenuse is the distance between the two cars. Using the pthagorean theorem we find that the distance is: + 0 = 6 miles. (b) To find the angle between Car A and the -ais, we draw the triangle made b dropping a vertical line from Car A s ending position. This triangle has side lengths 8 and 0 as the adjacent side and opposite side respectivel to the angle θ between Car A and the -ais. Thus we set up the equation: tan(θ) = 0 8 and solve this using arctan: θ = arctan( 0 8 ). The angle between the two cars is the difference between the angle Car B makes with the -ais and the angle Car A makes with the -ais. To find the angle Car B makes with the -ais, we will find the angle it makes with the -ais and then add. The angle, θ, Car B makes with the -ais comes from the triangle with side lengths 0 and. So tan θ = 0, and so θ = arctan ( ) 0. Now we add to find the angle Car B makes with the -ais and get arctan ( ) 0 +. Now we subtract the angle Car A makes with the -ais from the angle Car B makes with the -ais and get that the angle between the two cars is: ( ) arctan + 0 arctan(0 8 ). 7. A rope hanging straight down from a pole is feet longer than the pole. When it s stretched taut, it hits the ground 8 feet awa from the pole. How tall is the pole? Draw a picture of the situation. Call the length of the pole. Then we have a triangle with legs and 8, with hpotenuse +. Using the pthagorean theorem we have Now we can solve this equation: + 8 = ( + ) + 6 = = = + 6 = 7. A plane is fling at an altitude of 6,000 feet. Looking down, the pilot sees a ship at an angle of depression of 0 degrees. He sees a submarine at an angle of depression of 60 degrees in the same direction. How far apart are the ship and submarine? Page 8 of

19 Eam Review Draw a picture of the situation. We get two right triangles, Triangle A with the angle 0 with opposite side 6,000 and Triangle B with the angle 60 with opposite side 6,000. To find the distance between the ship and submarine we need find the difference between the adjacent side lengths of these two triangles. For Triangle A, call the adjacent side a. Then we have tan(0 ) = 6000 a = 6000 a a = = a For Triangle B, call the adjacent side b. Then we have tan(60 ) = 6000 b 6000 = b = b Now the distance between the ship and submarine is a b which is: = b = 60000( ) feet Here are a few additional problems that didn t make it into an of the above sections. 7. Calculate sin( ) without using a calculator. Use the difference identit for sine: ( sin ) ( ) = sin = = 6 ( ) cos ( ) ( ) cos sin 75. Simplif the epression using identities as needed: ( cos + cos ). It helps to first factor this equation: cos + cos = ( cos ) Then we see that we can appl the pthagorean identit sin = cos to obtain: ( cos ) = (sin ) = sin. Page 9 of

20 Eam Review 76. Calculate cos( 7 8 ) without using a calculator. Use the half angle formula for cosine: cos ( 7 8 ) = + cos( 7 8 ) = + cos( 7 ) = + = + Now, taking the square root of both sides we get ( ) 7 + cos = ± 8 but since the angle 7 8 be negative. Thus, is in the second quadrant, we know that the cosine of this angle should cos ( ) 7 + = Simplif tan +. Use the pthagorean identit tan + = sec and sec = cos and we see that tan + = sec = cos. 78. Rewrite the product as a sum or difference: sin(7t) cos(t). 79. True or false? cos + sin tan = sec. True. We need to start with cos + sin tan and transform it into sec. We use a series of trig identities: Page 0 of

21 Eam Review cos + sin tan = cos + sin sin cos = cos + sin cos = cos + sin cos = (cos + sin ) cos = () cos = cos = sec (pthagorean identit) 80. True or false? cot csc = cos (sec ) True. We need to start with cot csc identities: and transform it into cos (sec ). We use a series of trig cot csc ( ) cot = cot csc ( ) cot = (sec ) (pthagorean identit) csc ( cos ) = (sec ) sin sin switching to sine and cosine = cos (sec ) 8. Show work to prove that sec θ cos θ = sec θ. We need to start with identities: sec θ cos θ and transform it into sec θ. We use a series of trig sec θ cos θ = sec θ cos θ cos θ = cos θ cos θ cos θ = cos θ = cos θ = sec θ Page of